Решите уравнение: 2cos квадрате x-cosx-1=0

$2cos ^{2} -cosx-1=0$
$cosx=t$
$2t ^{2} -t-1=0$
D = (-1)² - 4· 2·(-1) = 1 + 8 = 9
$sqrt{D} =3$

$t_{1} = frac{1+3}{4} = frac{4}{4}=1$

$t _{2} = frac{1-3}{4} = frac{-2}{4}=- frac{1}{2}$

$left { {{cosx=1} atop {cosx=- frac{1}{2} }} right.$

$x=2 pi k,$ k∈Z
$x= frac{2 pi }{3}+2 pi n,$ n∈Z